\(a,n_{Br_2}=\dfrac{6,4}{160}=0,04\left(mol\right)\)

PTHH: C₂H₄ + Br₂ ---> C₂H₄Br₂

           0,04<---0,04

\(\rightarrow\left\{{}\begin{matrix}V_{C_2H_4}=0,04.22,4=0,896\left(l\right)\\V_{CH_4}=2,24-0,896=1,344\left(l\right)\end{matrix}\right.\\ b,\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,896}{2,24}.100\%=40\%\\\%V_{CH_4}=100\%-40\%=60\%\end{matrix}\right.\)

C2H4 + Br2 -> C2H4Br2

  a          a            a

C2H2 + 2Br2 -> C2H2Br4

   b          2b             b

n hỗn hợp khí = \(\dfrac{7.84}{22.4}=0.35mol\)

Ta có: \(\left\{{}\begin{matrix}a+b=0.35\\160a+320b=72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.25mol\\b=0.1mol\end{matrix}\right.\)

\(\%VC2H4=\dfrac{0.25\times22.4\times100}{7.84}=71.43\%\)

%VC2H2 = 100 - 71.43 = 28.57%

C2H4 + Br2 -> C2H4Br2

  a          a            a

C2H2 + 2Br2 -> C2H2Br4

   b          2b             b

n hỗn hợp khí = %VC2H4=0.25×22.4×1007.84=71.43%%VC2H4=0.25×22.4×1007.84=71.43%

%VC2H2 = 100 - 71.43 = 28.57%

Ai giúp e với ạ

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

a) Ta có: \(n_{C_2H_4}=\dfrac{9,1}{28}=0,325\left(mol\right)=n_{Br_2}\) \(\Rightarrow V_{Br_2}=\dfrac{0,325}{2}=0,1625\left(l\right)=162,5\left(ml\right)\)

b) Ta có: \(\left\{{}\begin{matrix}n_{C_2H_4}=0,325\left(mol\right)\\n_{CH_4}=\dfrac{13,44}{22,4}-0,325=0,275\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{hh}=9,1+0,275\cdot16=13,5\left(g\right)\)

c) PTHH: \(CH_4+2O_2 \underrightarrow{t^o} CO_2+2H_2O\)

                \(C_2H_4+3O_2 \underrightarrow{t^o} 2CO_2+ 2H_2O\)

Theo các PTHH: \(\Sigma n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=1,525\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,525\cdot22,4=34,16\left(l\right)\)

\(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right);n_{hh}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

            0,025<-0,125

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100\%=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)

a, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

_____0,15____0,3 (mol)

\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,15.22,4}{11,2}.100\%=30\%\)

\(\Rightarrow\%V_{CH_4}=100-30=70\%\)

b, - Khí thoát ra ngoài là CH₄.

\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)

Đáp án A

Đặt n_CH4 = x mol, n_C2H2 = y mol → x + y =0,25 mol

n_Br2(pư) = 2y =0,1 mol → y =0,05 mol

→ x = 0,2 mol

→ V_CH4 = 4,48 lít và V_C2H2 =1,12 lít

\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)

\(\Rightarrow n_{C_2H_4}=0,05mol\Rightarrow m_{C_2H_4}=1,4g\)

\(\%m_{C_2H_4}=\dfrac{1,4}{2}\cdot100\%=70\%\)

\(\%m_{CH_4}=100\%-70\%=30\%\)

Đáp án C

C₂H₅Br -> C₂H₄ + HBr ( xúc tác ancol)

C₂H₄ + Br₂ -> C₂H₄Br₂

n_C2H5Br = n_Br2 = 0,05mol

=> m = 5,45g